Simplify; express your answer in exponential form. Assume $z\neq 0, r\neq 0$. $\dfrac{{(z)^{4}}}{{(z^{2}r^{-1})^{-2}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${z}$ to the exponent ${4}$ . Now ${1 \times 4 = 4}$ , so ${(z)^{4} = z^{4}}$ In the denominator, we can use the distributive property of exponents. ${(z^{2}r^{-1})^{-2} = (z^{2})^{-2}(r^{-1})^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(z)^{4}}}{{(z^{2}r^{-1})^{-2}}} = \dfrac{{z^{4}}}{{z^{-4}r^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{z^{4}}}{{z^{-4}r^{2}}} = \dfrac{{z^{4}}}{{z^{-4}}} \cdot \dfrac{{1}}{{r^{2}}} = z^{{4} - {(-4)}} \cdot r^{- {2}} = z^{8}r^{-2}$.